Recent content by rickyjoepr

got it Fk*d = ((.5)mVi^2) + (.5)((Iw^2) eventually simplified to Sqrt [ (4/r^2)* (.42*10.6*1280 - (1/2)Vi^2) ] = w

Homework Statement coefficient of kinetic friction between the disk and the surface is 0.42 gravity = 10.6 m/s disk mass = 1.75x10^9 the skid marks are 1280m long, This is due to the fact that uneven friction had set the saucer in very slow rotation around its principal axis. By...

Homework Statement A saucer of mass 1.75x10^9 enters 1.the gravitational field of PlanetX and enters its atmosphere; the gravitational field of the planet is 10.6 m/s2 and is considered uniform throughout the 80-km thick atmosphere. If the saucer enters the atmosphere at 460 m/s and slows...

Everything worked out, I had been at that for hours, thank you for your guidance. I learned something

I thought it was an issue with G, I was trying to convert G to km , but I will try what you suggested edit: and even if i dd convert G to km, it would still be wrong as the kg Mass of Earth must work with m/s

Homework Statement an asteroid heading towards earth, has a speed of 7.25 km/s as it passes the moons orbit, kit misses Earth by 5000km. What is the speed at the closest point to earth? The professor provided us with the solution, which should be 11.0 km/s, however when I do the calculation...